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Let $X$ be a del Pezzo surface (over $\mathbb{C}$), which is obtained by a blow up $\pi: X \rightarrow \mathbb{P}^{2}$ in a collection of points. Let $H$ be the hyperplane class of $\mathbb{P}^{2}$.

Question: Consider smooth, irreducible curves $\Sigma$ satisfying the inequality $-K_{X}\cdot \Sigma > \frac{1}{2}\Sigma \cdot \Sigma + \pi^{*}H \cdot \Sigma $.

Is the genus of such curves bounded above?

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Here is one proof that the genus is bounded (although presumably a more elegant solution exists).

Write $\Sigma=d\pi^*H - \sum_{i=1}^nm_iE_i$ in the usual basis for $\operatorname{NS}(X)$ (i.e. $\Sigma$ is the strict transform of a curve of degree $d$ in $\mathbb{P}^2$, with multiplicity $m_i$ at each of the points which is blown up).

We can rearrange the inequality into the form $\tfrac12(\Sigma^2 + K_X\Sigma) < \tfrac12(d-\sum m_i)$. Then, from the usual genus formula for $\Sigma\subset X$, we have \begin{equation*} g_\Sigma = 1 + \tfrac12(\Sigma^2 + K_X\Sigma) < 1 + \tfrac12\left(d-\sum_{i=1}^n m_i\right) \end{equation*} so we may assume that $d > 2 + \sum m_i$, else $g_\Sigma \leq 2$.

Now we can prove the result by induction on $n$ (the number of points blown up on $\mathbb{P}^2$). If $n=0$ it is easy to see that the inequality becomes $3d > \tfrac12d^2 + d$, hence $d<4$ and $g\leq1$.

By the induction hypothesis, if $n>0$ then we may assume each $m_i\geq 1$ (else blow down the point with $m_i=0$ and reduce to the $n-1$ case). The inequality is \begin{equation*} 3d - \sum_{i=1}^nm_i > \frac12 \left(d^2 - \sum_{i=1}^n m_i^2 \right) + d \end{equation*} which can be rearranged as \begin{equation*} (d - 2)^2 < \sum_{i=1}^n (m_i-1)^2 + 4 - n \end{equation*} But, since $m_i\geq 1 \: \forall i$, we have $\sum (m_i-1)^2 \leq \left(\sum(m_i-1)\right)^2$ and since $\sum m_i < d - 2$ we have \begin{equation*} (d - 2)^2 < (d-2-n)^2 + 4 - n \end{equation*} which gives the bound $d < \frac{n^2+3n+4}{2n}$, and hence a bound for $g$.

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  • $\begingroup$ Very nice, thanks Tom! I think the inequality coming after "The inequality is..." should be reversed, (although the sign is correct again by the next line, so it doesn't affect anything). $\endgroup$ – Nick L Jan 31 '18 at 19:33

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